Roots of $e^{(z^{2})}=1$

Question

I’m looking for complex roots of $e^{(z^{2})}=1$ on the inside of the circle |z|=3. I have found $0, \sqrt{\pi}\cdot \{1+i,1-i,-1+i,-1-i\}$. Is that it, wolfram alpha doesn’t help:/

Thank you

Answer 1

Your answer is correct.

Let $z = x + iy$, so we have: $$ e^{z^2}=1 \iff e^{x^2-y^2}e^{2ixy} = 1 \iff \left\{ \begin{align}x^2-y^2&=0,\\ 2xy &= 2k\pi,\ k\in\mathbb Z.\end{align}\right. $$

Thus, considering positive and negative $k$'s, we end up with cases where $x = y$ or $x = - y$, so $z = \sqrt{k\pi}(\pm 1 \pm i),\ k\in\mathbb N$.

Finally, $|z|\leq 3 \iff \sqrt{2k\pi} \leq 3 \iff k \leq \frac 9{2\pi}\iff k = 0,1.$ These give the 5 solutions that you listed.

Here is a Wolfram Alpha link for the solutions.